1. Suppose you are in charge of inventory maintenance at a bicycle shop. One of

Question

1. Suppose you are in charge of inventory maintenance at a bicycle shop. One of your jobs is to ensure that the tire pressure in cach of the display bicycles is between 65-85 PSI (pounds per square inch). If the pressure is too low, then there is a risk of wheel damage when a customer rides out on one. On the other end, if the pressure is too high, there is a (small) risk of the tire exploding! Of course, you don't know the true pressure of any particular tire. Instead, you have the output from your pressure gauge. This wi be similar to the true pressure, but not nccessarily the same. Laboratory testing of the particular gauge you use has shown that there is a2 PSI error margin and so to be careful you decide that vou will adjust the pressure on any tire that has a measured pressure above 81 PSI or below 69 PSI, giving you 2x the error margin on either side. Previous testing shows that this procedure will give the following results: Mcasure inside 69-81PSI Mcasure outside 69-81 PSI PSI within 65-85 PSI PSI outside 65-85 PSI 5000 105 216 (a) State (in words) the appropriate nul and alternate hypothesis. (b) Calculate the value of a. What situation does the Type I error rate represent here? (c) Calculate the value of B. What situation does the Type II error rate represent here? d) Calculate the Power of this test. What situation does test Power represent here? (e) What is the rejection region for the test you are conducting? (f) Practically speaking, which error seems more problematic: a Type I error or a Type II error? (g If we wanted to decrease the Type I error rate, should we increase or decrease the rejection region? (h) If we wanted to decrcase the Type II error rate, should we increase or decrease the rejection region?

Explanation / Answer

Answer to the question)

Part a)

Null hypothesis: The gauge provides the true pressure

Alternate hypothesis: Gauge doesnot provide the true pressure value

.

part b)

Type I error is that we incorrectly reject the null

the true value is within 65 to 85 , but the guage reads it wrong

so despite of the null being true we end up rejecting it

Alpha = (105 ) / (5000+105) = 0.0206

.

Answer to part c)

Type II error is that the null is wrong , the reading is outside 65 to 85 , but the gauge shows it inside

Beta = 3 /(216+3) = 0.01370

.

Answer to part d)

Power of test = 1 - Beta

Power of test = 1 - 0.01370 = 0.9863

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